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- 1. Electric Circuit Analysis Basic analysis techniques Ahsan Khawaja Ahsan_khawaja@comsats.edu.pk Lecturer Room 102 Department of Electrical Engineering
- 2. Circuit Abstraction Question: What is the current through the bulb? Concept of Abstraction Solution: In order to calculate the current, we can replace the bulb with a resistor. R is the only subject of interest, which serves as an abstraction of the bulb.
- 3. Voltage (Potential) a b VVab 5 a、b, which point’s potential is higher？ b a V6a V V4b V Vab = ? a b +Q from point b to point a get energy ，Point a is Positive? or negative ? Basic Quantities Example
- 4. Voltage (Potential) ab c´ c d d´ 2211 21 221121222 2 21112 1111 111 1b1bb 0 )( )( , 0 rRrR EE I rRrRIEEIrEVIrVV EVV RrRIEIRVV rRIEIrVV IREVEV IRVIRVVVV V dda dd cd cc bc aab a Basic Quantities Example I
- 5. Open Circuit R= I=0, V=E , P=0 E R0 Short Circuit R=0 E R0 R＝0 0 R E I 00 IREV 0 2 RIPE Basic Quantities
- 6. Nodes, Branches, Loops, mesh • A circuit containing three nodes and five branches. • Node 1 is redrawn to look like two nodes; it is still one nodes. Kirchhoff's Current and Voltage Laws
- 7. • sum of all currents entering a node is zero • sum of currents entering node is equal to sum of currents leaving node KCL KCL Mathematically i1(t) i2(t) i4(t) i5(t) i3(t) n j j ti 1 0)( n j jI 1 0 Kirchhoff's Current and Voltage Laws
- 8. • sum of all currents entering a node is zero • sum of currents entering node is equal to sum of currents leaving node KCL P1.9 DCBA iiii Kirchhoff's Current and Voltage Laws In Out 0A B C O I I i i i i
- 9. KCL + -120V 50* 1W Bulbs Is • Find currents through each light bulb: IB = 1W/120V = 8.3mA • Apply KCL to the top node: IS - 50IB = 0 • Solve for IS: IS = 50 IB = 417mA KCL-Christmas Lights Kirchhoff's Current and Voltage Laws
- 10. KCL Current divider N V G1 G2 I + - I1 I2 121 21 11 1 11 RRR RR I R RI R V I I RR R I 21 1 2 Kirchhoff's Current and Voltage Laws In case of parallel : 1 2 1 2 1 1 1 , , V= I I G G G R R R R G
- 11. sum of voltages around any loop in a circuit is zero. KVL • A voltage encountered + to - is positive. • A voltage encountered - to + is negative. KVL Mathematically 0)( 1 n j j tv 0 1 n j jV Kirchhoff's Current and Voltage Laws
- 12. KVL Determine the voltages Vae and Vec. Kirchhoff's Current and Voltage Laws 10 24 0ae V 16 12 4 6 0ae V 4 + 6 + Vec = 0
- 13. KVL Voltage divider R1 R2 - V1 + + - V2 + - V 21 1 11 RR R VIRV 21 2 22 RR R VIRV Important voltage Divider equations N Kirchhoff's Current and Voltage Laws
- 14. KVL Voltage divider kR 151 Volume control? Example: Vs = 9V, R1 = 90kΩ, R2 = 30kΩ Kirchhoff's Current and Voltage Laws
- 15. Mesh Analysis • Technique to find voltage drops around a loop using the currents that flow within the loop, Kirchoff’s Voltage Law, and Ohm’s Law – First result is the calculation of the mesh currents • Which can be used to calculate the current flowing through each component – Second result is a calculation of the voltages across the components • Which can be used to calculate the voltage at the nodes.
- 16. Mesh Analysis What is a mesh: It is a loop which does not contain any other loops within it. i1 i2 This circuit has two meshes
- 17. Steps in Mesh Analysis Vin
- 18. Step 1 • Identify all of the meshes in the circuit Vin
- 19. Step 2 • Label the currents flowing in each mesh i1 i2 Vin
- 20. Step 3 • Label the voltage across each component in the circuit i1 i2 + V1 _ Vin + V3 _ + V5 _ + V6 _ + V2 - + V4 -
- 21. Step 4 • Use Kirchoff’s Voltage Law i1 i2 + V1 _ Vin + V3 _ + V5 _ + V6 _ + V2 - + V4 - 0 0 543 6321 VVV VVVVVin
- 22. Step 5 • Use Ohm’s Law to relate the voltage drops across each component to the sum of the currents flowing through them. – Follow the sign convention on the resistor’s voltage. RIIV baR
- 23. Step 5 i1 i2 + V1 _ Vin + V3 _ + V5 _ + V6 _ + V2 - + V4 - 616 525 424 3213 212 111 RiV RiV RiV RiiV RiV RiV
- 24. Step 6 • Solve for the mesh currents, i1 and i2 – These currents are related to the currents found during the nodal analysis. 213 542 62171 iiI IIi IIIIi
- 25. Step 7 • Once the mesh currents are known, calculate the voltage across all of the components.
- 26. 12V
- 27. From Previous Slides 616 525 424 3213 212 111 RiV RiV RiV RiiV RiV RiV 0 0 543 6321 VVV VVVVVin
- 28. Substituting in Numbers kiV kiV kiV kiiV kiV kiV 1 3 6 5 8 4 16 25 24 213 12 11 0 012 543 6321 VVV VVVVV
- 29. Substituting the results from Ohm’s Law into the KVL equations 0365 0158412 2221 12111 kikikii kikiikikiV
- 30. Results • One or more of the mesh currents may have a negative sign. Mesh Currents ( A) i1 740 i2 264
- 31. Results Currents ( A) IR1 = i1 740 IR2 = i1 740 IR3 = i1- i2 476 IR4 = i2 264 IR5 = i2 264 IR6 = i1 740 IVin = i1 740 The currents through each component in the circuit.
- 32. Example Using Mesh analysis find the currents through each resistor KVL in loop 1: -2+2i1+4(i1-i2)= 0 6i1-4i2-2= 0 6i1-4i2= 2 …………..1) KVL in loop 2: 6+4(i2-i1)+i2= 0 -4i1+5i2= -6 ………..2) From eqn 1)&2) Loop currents: i1= -1A, i2= -2A Currents through each resistor: i2 = i1= -1A i1 = i2 = -2A i4 = i1-i2 = -1A-(-2A) = 1A i1 i2
- 33. Example Use the mesh current method to determine the power associated with each voltage source in the circuit. Calculate the voltage vo across the 8 resistor. KVL in Loop 1: -40+2i1+8(i1-i2)=0 KVL in Loop 2: 8(i2-i1)+6i2+6(i2-i3)=0 KVL in Loop 3: 6(i3-i2)+4i3+20=0 i1 i2 i3
- 34. 10i1-8i2+0i3=40 -8i1+20i2-6i3=0 0i1-6i2+10i3= -20 By solving the matrix i1=5.6A i2= 2A i3= -0.8A P40v= -40i1= -40(5.6) = -224W P20v = 20i3= 20(-0.8) = -16W vo= 8(i1-i2)= 8(5.6-2) = 8(3.6)=28.8V
- 35. 35 Example Apply KVL to each mesh 2 1 7 5 0s V v v v 2 6 7 0v v v 15 3 0s v v v Mesh 1: Mesh 2: Mesh 3: 14 8 6 0s v v V vMesh 4: DC DC 1R 3R 5R 7R 2R 6R 8R 4R 1v 2v 3v 4v 5v 6v 7v 8v + + + + + + + + - - - - - - - - 1sV 2sV 1i 2i 3i 4i
- 36. 36 DC DC 1R 3R 5R 7R 2R 6R 8R 4R 1v 2v 3v 4v 5v 6v 7v 8v + + + + + + + + - - - - - - - - 1sV 2sV 1i 2i 3i 4i 2 1 7 5 0s V v v v 2 6 7 0v v v 15 3 0s v v v Mesh 1: Mesh 2: Mesh 3: 14 8 6 0s v v V vMesh 4: 2 1 1 1 2 7 1 3 5 ( ) ( ) 0s V i R i i R i i R 2 2 2 4 6 2 1 7 ( ) ( ) 0i R i i R i i R 13 1 5 3 3 ( ) 0s i i R V i R Mesh 1: Mesh 2: Mesh 3: 14 4 4 8 4 2 6 ( ) 0s i R i R V i i RMesh 4: Express the voltage in terms of the mesh currents:
- 37. 37 2 1 1 1 2 7 1 3 5 ( ) ( ) 0s V i R i i R i i R 2 2 2 4 6 2 1 7 ( ) ( ) 0i R i i R i i R 13 1 5 3 3 ( ) 0s i i R V i R Mesh 1: Mesh 2: Mesh 3: 14 4 4 8 4 2 6 ( ) 0s i R i R V i i RMesh 4: Mesh 1: Mesh 2: Mesh 3: Mesh 4: 21 5 7 1 7 2 5 3 ( ) s R R R i R i R i V 7 1 2 6 7 2 6 4 ( ) 0R i R R R i R i 15 1 3 5 3 ( ) s R i R R i V 16 2 4 6 8 4 ( ) s R i R R R i V
- 38. 38 Mesh 1: Mesh 2: Mesh 3: Mesh 4: 21 5 7 1 7 2 5 3 ( ) s R R R i R i R i V 7 1 2 6 7 2 6 4 ( ) 0R i R R R i R i 15 1 3 5 3 ( ) s R i R R i V 16 2 4 6 8 4 ( ) s R i R R R i V 2 1 1 1 5 7 7 5 1 7 2 6 7 6 2 5 3 5 3 6 4 6 8 4 0 00 0 0 0 0 s s s VR R R R R i R R R R R i VR R R i R R R R i V

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